Open In App

Print ancestors of a given binary tree node without recursion

Last Updated : 24 Apr, 2025
Comments
Improve
Suggest changes
Like Article
Like
Report

Given a Binary Tree and a node key. The task is to print all the ancestors of the given key in the tree without using recursion. An ancestor of a node is any node that lies on the path from the root to that node, excluding the node itself.
Note: All node values are unique, and the given key is guaranteed to be present in the tree.

Examples:

Input: key = 7

1
/ \
2 3
/ \
4 5
/ \
7 8
Output: 5 2 1
Explanation: The path from the root (1) to 7 is 1 -> 2 -> 5 -> 7, so its ancestors are 5, 2, and 1.

Input: key = 50

10
/ \
20 30
/ \
40 50
Output: 20 10
Explanation: The path from the root (10) to 50 is 10 -> 20 -> 50, so its ancestors are 20 and 10.

A Recursive Solution for this problem is discussed here. 

Approach:

The idea is to find the ancestors of a node in a binary tree without using recursion. The approach uses an iterative Depth-First Search (DFS) with a stack to traverse the tree, and a parent mapping to store each node’s parent. Once the node with the key is found, the ancestors are collected by following the parent pointers.

Steps to implement the above idea:

  • Create a Node class to represent each tree node with data, left, and right pointers.
  • Initialize a stack for DFS traversal and an unordered map called parent to store the parent-child relationships.
  • Traverse the tree iteratively using the stack, pushing nodes and their corresponding parent into the map.
  • Once the node with the given key is found, start collecting ancestors by tracing the parent pointers.
  • Finally, print the collected ancestors after the parent mapping is complete.
C++ 
// C++ Code to print all ancestors of a given key
// in a Binary Tree without Recursion
#include <bits/stdc++.h>
using namespace std;

class Node {
public:
    int data;
    Node* left;
    Node* right;

    Node(int x) {
        data = x;
        left = right = nullptr;
    }
};

// Function to find and return ancestors of 
// the given key
vector<int> printAncestors(Node* root, int key) {
    
    if (!root) {
        return {};
    }
    
    unordered_map<Node*, Node*> parent;
    stack<Node*> stk;
    stk.push(root);
    
    // Build parent mapping using iterative DFS
    while (!stk.empty()) {
        Node* curr = stk.top();
        stk.pop();
        
        if (curr->right) {
            parent[curr->right] = curr;
            stk.push(curr->right);
        }
        if (curr->left) {
            parent[curr->left] = curr;
            stk.push(curr->left);
        }
    }
    
    vector<int> ancestors;
    Node* node = nullptr;
    stk.push(root);
    
    // Find the node with the given key
    while (!stk.empty()) {
        Node* curr = stk.top();
        stk.pop();
        
        if (curr->data == key) {
            node = curr;
            break;
        }
        
        if (curr->right) {
            stk.push(curr->right);
        }
        if (curr->left) {
            stk.push(curr->left);
        }
    }
    
    // Collect ancestors
    while (parent.find(node) != parent.end()) {
        node = parent[node];
        ancestors.push_back(node->data);
    }
    
    return ancestors;
}

void printArray(vector<int>& arr) {
    for (int num : arr) {
        cout << num << " ";
    }
    cout << endl;
}

int main() {
    
    // Constructing the binary tree
    //        1
    //       / \
    //      2   3
    //     / \
    //    4   5
    //       / \
    //      7   8
    Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->left->right = new Node(5);
    root->left->right->left = new Node(7);
    root->left->right->right = new Node(8);

    int key = 7;
    vector<int> result = printAncestors(root, key);
    
    printArray(result); 
   
    return 0;
}
C 
// C Code to print all ancestors of a given key
// in a Binary Tree without Recursion
#include <stdio.h>
#include <stdlib.h>

struct Node {
    int data;
    struct Node* left;
    struct Node* right;
};

// Function to create a new node
struct Node* createNode(int x) {
    struct Node* newNode 
      = (struct Node*)malloc(sizeof(struct Node));
    newNode->data = x;
    newNode->left = NULL;
    newNode->right = NULL;
    return newNode;
}

// Function to print the array of ancestors
void printArray(int arr[], int size) {
    for (int i = 0; i < size; i++) {
        printf("%d ", arr[i]);
    }
    printf("\n");
}

// Function to find and return ancestors of the given key
void printAncestors(struct Node* root, int key) {
    
    if (!root) {
        return;
    }
    
    struct Node* parent[1000] = {NULL};
    struct Node* stack[1000];
    int top = -1;
    
    stack[++top] = root;
    
    // Build parent mapping using iterative DFS
    while (top != -1) {
        struct Node* curr = stack[top--];
        
        if (curr->right) {
            parent[curr->right->data] = curr;
            stack[++top] = curr->right;
        }
        if (curr->left) {
            parent[curr->left->data] = curr;
            stack[++top] = curr->left;
        }
    }
    
    struct Node* node = NULL;
    stack[++top] = root;
    
    // Find the node with the given key
    while (top != -1) {
        struct Node* curr = stack[top--];
        
        if (curr->data == key) {
            node = curr;
            break;
        }
        
        if (curr->right) {
            stack[++top] = curr->right;
        }
        if (curr->left) {
            stack[++top] = curr->left;
        }
    }
    
    // Collect ancestors
    int ancestors[1000];
    int count = 0;
    
    while (node && parent[node->data] != NULL) {
        node = parent[node->data];
        ancestors[count++] = node->data;
    }
    
    // Print ancestors
    printArray(ancestors, count);
}

int main() {
    
    // Constructing the binary tree
    //        1
    //       / \
    //      2   3
    //     / \
    //    4   5
    //       / \
    //      7   8
    struct Node* root = createNode(1);
    root->left = createNode(2);
    root->right = createNode(3);
    root->left->left = createNode(4);
    root->left->right = createNode(5);
    root->left->right->left = createNode(7);
    root->left->right->right = createNode(8);

    int key = 7;
    printAncestors(root, key);
   
    return 0;
}
Java 
// Java Code to print all ancestors of a given key
// in a Binary Tree without Recursion
import java.util.*;

class Node {
    int data;
    Node left, right;

    Node(int x) {
        data = x;
        left = right = null;
    }
}

class GfG {
    
    // Function to find and return ancestors of 
    // the given key
    static int[] printAncestors(Node root, int key) {
        
        if (root == null) {
            return new int[0];
        }
        
        Map<Node, Node> parent = new HashMap<>();
        Stack<Node> stk = new Stack<>();
        stk.push(root);
        
        // Build parent mapping using iterative DFS
        while (!stk.isEmpty()) {
            Node curr = stk.pop();
            
            if (curr.right != null) {
                parent.put(curr.right, curr);
                stk.push(curr.right);
            }
            if (curr.left != null) {
                parent.put(curr.left, curr);
                stk.push(curr.left);
            }
        }
        
        List<Integer> ancestors = new ArrayList<>();
        Node node = null;
        stk.push(root);
        
        // Find the node with the given key
        while (!stk.isEmpty()) {
            Node curr = stk.pop();
            
            if (curr.data == key) {
                node = curr;
                break;
            }
            
            if (curr.right != null) {
                stk.push(curr.right);
            }
            if (curr.left != null) {
                stk.push(curr.left);
            }
        }
        
        // Collect ancestors
        while (parent.containsKey(node)) {
            node = parent.get(node);
            ancestors.add(node.data);
        }
        
        return ancestors.stream().mapToInt(i -> i).toArray();
    }

    static void printArray(int[] arr) {
        for (int num : arr) {
            System.out.print(num + " ");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        
        // Constructing the binary tree
        //        1
        //       / \
        //      2   3
        //     / \
        //    4   5
        //       / \
        //      7   8
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.left.right.left = new Node(7);
        root.left.right.right = new Node(8);

        int key = 7;
        int[] result = printAncestors(root, key);
        
        printArray(result); 
    }
}
Python 
# Python Code to print all ancestors of a given key
# in a Binary Tree without Recursion

class Node:
    def __init__(self, x):
        self.data = x
        self.left = None
        self.right = None

# Function to find and return ancestors
# of the given key
def printAncestors(root, key):
    
    if not root:
        return []
    
    parent = {}
    stk = [root]
    
    # Build parent mapping using iterative DFS
    while stk:
        curr = stk.pop()
        
        if curr.right:
            parent[curr.right] = curr
            stk.append(curr.right)
        if curr.left:
            parent[curr.left] = curr
            stk.append(curr.left)

    ancestors = []
    node = None
    stk.append(root)
    
    # Find the node with the given key
    while stk:
        curr = stk.pop()
        
        if curr.data == key:
            node = curr
            break
        
        if curr.right:
            stk.append(curr.right)
        if curr.left:
            stk.append(curr.left)

    # Collect ancestors
    while node in parent:
        node = parent[node]
        ancestors.append(node.data)

    return ancestors

def printArray(arr):
    print(" ".join(map(str, arr)))

if __name__ == "__main__":
    
    # Constructing the binary tree
    #        1
    #       / \
    #      2   3
    #     / \
    #    4   5
    #       / \
    #      7   8
    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.left.right = Node(5)
    root.left.right.left = Node(7)
    root.left.right.right = Node(8)

    key = 7
    result = printAncestors(root, key)
    
    printArray(result)
C# 
// C# Code to print all ancestors of a given key
// in a Binary Tree without Recursion
using System;
using System.Collections.Generic;

class Node {
    public int data;
    public Node left, right;

    public Node(int x) {
        data = x;
        left = right = null;
    }
}

class GfG {
    
    // Function to find and return ancestors of 
    // the given key
    public static int[] printAncestors(Node root, int key) {
        
        if (root == null) {
            return new int[0];
        }
        
        Dictionary<Node, Node> parent = new Dictionary<Node, Node>();
        Stack<Node> stk = new Stack<Node>();
        stk.Push(root);
        
        // Build parent mapping using iterative DFS
        while (stk.Count > 0) {
            Node curr = stk.Pop();
            
            if (curr.right != null) {
                parent[curr.right] = curr;
                stk.Push(curr.right);
            }
            if (curr.left != null) {
                parent[curr.left] = curr;
                stk.Push(curr.left);
            }
        }
        
        List<int> ancestors = new List<int>();
        Node node = null;
        stk.Push(root);
        
        // Find the node with the given key
        while (stk.Count > 0) {
            Node curr = stk.Pop();
            
            if (curr.data == key) {
                node = curr;
                break;
            }
            
            if (curr.right != null) {
                stk.Push(curr.right);
            }
            if (curr.left != null) {
                stk.Push(curr.left);
            }
        }
        
        // Collect ancestors
        while (parent.ContainsKey(node)) {
            node = parent[node];
            ancestors.Add(node.data);
        }
        
        return ancestors.ToArray();
    }

    public static void printArray(int[] arr) {
        foreach (int num in arr) {
            Console.Write(num + " ");
        }
        Console.WriteLine();
    }

    public static void Main() {
        
        // Constructing the binary tree
        //        1
        //       / \
        //      2   3
        //     / \
        //    4   5
        //       / \
        //      7   8
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.left.right.left = new Node(7);
        root.left.right.right = new Node(8);

        int key = 7;
        int[] result = printAncestors(root, key);
        
        printArray(result); 
    }
}
JavaScript 
// JavaScript Code to print all ancestors of a given key
// in a Binary Tree without Recursion

class Node {
    constructor(x) {
        this.data = x;
        this.left = null;
        this.right = null;
    }
}

// Function to find and return ancestors of 
// the given key
function printAncestors(root, key) {
    
    if (!root) {
        return [];
    }
    
    let parent = new Map();
    let stk = [];
    stk.push(root);
    
    // Build parent mapping using iterative DFS
    while (stk.length > 0) {
        let curr = stk.pop();
        
        if (curr.right) {
            parent.set(curr.right, curr);
            stk.push(curr.right);
        }
        if (curr.left) {
            parent.set(curr.left, curr);
            stk.push(curr.left);
        }
    }
    
    let ancestors = [];
    let node = null;
    stk.push(root);
    
    // Find the node with the given key
    while (stk.length > 0) {
        let curr = stk.pop();
        
        if (curr.data === key) {
            node = curr;
            break;
        }
        
        if (curr.right) {
            stk.push(curr.right);
        }
        if (curr.left) {
            stk.push(curr.left);
        }
    }
    
    // Collect ancestors
    while (parent.has(node)) {
        node = parent.get(node);
        ancestors.push(node.data);
    }
    
    return ancestors;
}

function printArray(arr) {
    console.log(arr.join(" "));
}

// Constructing the binary tree
//        1
//       / \
//      2   3
//     / \
//    4   5
//       / \
//      7   8
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.left.right.left = new Node(7);
root.left.right.right = new Node(8);

let key = 7;
let result = printAncestors(root, key);

printArray(result);

Output
5 2 1

Time Complexity: O(n) – Tree traversal visits each node once.
Space Complexity: O(n) – Stack and parent map store nodes.


Next Article
Print ancestors of a given binary tree node without recursion

K
kartik
Improve
Article Tags :
Practice Tags :

Similar Reads

We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox